Elasticity and Strength of Materials: Study Notes

Elasticity and Strength of Materials

Introduction

This section explores the physical principles underlying the elasticity and strength of materials, with a focus on biological tissues such as bone. Key concepts include stress, strain, Young's modulus, energy absorption during fracture, and the physics of impulsive forces in collisions.

Stretch and Compression

• Stress (S): The internal force per unit area acting on a material. Defined as: $S = \frac{F}{A}$ where F is the applied force and A is the cross-sectional area.

• Strain (S_l): The fractional change in length due to applied force. For longitudinal strain: $S_l = \frac{\Delta l}{l}$ where $\Delta l$ is the change in length and $l$ is the original length.

• Hooke's Law: Within the elastic limit, the ratio of stress to strain is constant: $Y = \frac{S}{S_l}$ where Y is Young's modulus.

Young's Modulus

Young's modulus is a measure of the stiffness of a material. It quantifies the relationship between stress and strain in the linear (elastic) region of deformation.

Material	Young's modulus (dyn/cm2)	Rupture strength (dyn/cm2)
Steel	200 × 1011	450 × 107
Aluminum	69 × 1011	6.2 × 107
Bone	14 × 1010	100 × 106 (compression)
Tendon	27.5 × 108	1.2 × 106 (stretch)
Muscle	4.0 × 106	8.55 × 105 (stretch)

Additional info: Young's modulus is typically higher for stiffer materials such as steel and lower for biological tissues like muscle.

Spring

• Hooke's Law for Springs: The force required to stretch or compress a spring is proportional to the displacement: $F = -K \Delta l$ where K is the spring constant.

• Potential Energy in a Spring: $E = \frac{1}{2} K \Delta l^2$

• Elastic Body Analogy: An elastic body under stress behaves like a spring with an effective spring constant: $K = \frac{Y A}{l}$

• Potential Energy in an Elastic Body: $E = \frac{1}{2} \frac{Y A}{l} \Delta l^2$

Bone Fracture

• Energy to Break a Bone: The energy required to break a bone of area A and length l can be calculated assuming the bone remains elastic until fracture.

• Breaking Stress (S_B): The maximum stress a bone can withstand before fracturing.

• Fracture Force (F_B): $F_B = S_B A = \frac{Y A}{l} \Delta l$

• Compression at Breaking Point: $\Delta l = \frac{S_B l}{Y}$

• Energy Stored at Fracture: $E = \frac{1}{2} \frac{Y A}{l} (\Delta l)^2 = \frac{1}{2} A l \frac{S_B^2}{Y}$

Example Calculation

• Given: Two leg bones, combined length $l = 90$ cm, area $A = 6$ cm2, $S_B = 10^6$ dyn/cm2, $Y = 14 \times 10^{10}$ dyn/cm2.

• Energy absorbed by one leg at fracture: $E = \frac{1}{2} \frac{6 \times 90 \times 10^{18}}{14 \times 10^{10}} = 19.25 \times 10^8 \text{ erg} = 192.5 \text{ J}$

• Combined energy for two legs: $385$ J.

• This is the energy equivalent to a 70-kg person jumping from a height of 56 cm.

Impulsive Force

• Impulsive Force: A large force exerted over a short time interval during a collision.

• Exact magnitude is difficult to determine, but the average value can be calculated using momentum change:

$F_{av} = \frac{m v_f - m v_i}{\Delta t}$

Fracture Due to a Fall

• When a person falls from height $h$, the velocity on impact is: $v = \sqrt{2gh}$

• Momentum on the ground: $mv = m\sqrt{2gh}$

• Average impact force: $F = \frac{m}{\Delta t} \sqrt{2gh}$

Airbags

• Airbags expand rapidly in a collision to increase the stopping distance and reduce the force on the passenger.

• Average deceleration: $a = \frac{v^2}{2s}$ where $s$ is the stopping distance.

• Average force: $F = m a = m \frac{v^2}{2s}$

• For a 70-kg person with a 30-cm stopping distance: $F = \frac{70 \times 10^3 v^2}{2 \times 30} = 1.17 \times 10^3 v^2$ (dyn)

• If the force is distributed over 1000 cm2, the applied force per cm2 is $4.42 \times 10^3$ dyn, just below the estimated strength of body tissue.

Summary Table: Key Quantities and Formulas

Quantity	Formula	Description
Stress	$S = \frac{F}{A}$	Force per unit area
Strain	$S_l = \frac{\Delta l}{l}$	Fractional change in length
Young's modulus	$Y = \frac{S}{S_l}$	Stiffness of material
Hooke's Law (spring)	$F = -K \Delta l$	Force-displacement relation for springs
Potential energy (spring)	$E = \frac{1}{2} K \Delta l^2$	Energy stored in a spring
Impulsive force (average)	$F_{av} = \frac{m v_f - m v_i}{\Delta t}$	Average force during collision
Impact velocity (fall)	$v = \sqrt{2gh}$	Velocity on impact from height $h$
Average force (airbag)	$F = m \frac{v^2}{2s}$	Force with stopping distance $s$

Applications

• Understanding bone fracture mechanics helps in designing safer vehicles and protective gear.

• Airbags and crumple zones in cars utilize the principles of impulse and energy absorption to minimize injury.